[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 576
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There's enough to go around.
Or am I missing something?
The most important thing is to stop in time. How do you know if you've gotten to that particular unlit one? We can't make any notches.
I'm wrong, it'll just be the last one unlit when we get to the end. Seems like the right decision.
I don't fully understand it myself yet. But I didn't look for answers.
P.S. 2 MikeM: The answer is correct.
P.P.S. What if you stupidly light all the unlit, and then knock out all - now counting? Probably wrong, as surely the number of on/off attempts for each horse is limited to 1.
Good algorithmic problem, solved the same problem only with wagons on braingames.ru
There's clearly something missing from the condition. At least it doesn't make sense to me.
Can we have a condition about the wagons? Could there be more clarity there?
A lot of carriages are interlocked in a ring. The lights in the carriages are on in random order, i.e. somewhere they are on, somewhere they are off. You need to count the wagons. You can only turn on/off the lights in them.
Many carriages are linked together in a ring. The lights in the carriages are on in a random order, i.e. somewhere on, somewhere off. You have to count the wagons. You can only turn the lights on/off in them.
Uh-oh! That's better. So the merry-go-round is spinning and we're standing outside the fence. It remains to be seen how we can control the lanterns.
And the merry-go-round and the cars are stationary. You don't have to stand around and count. You have to walk around the carriages inside, and in the carousel from one horse to the next. The number of light switches is unlimited.
How is that not a solution?
Why isn't this the solution?
How do you leave one unlit? You can't see the whole picture.